gr. 11 physics
TRANSCRIPT
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Copyright 2011 Nelson Education Ltd. Unit 1: Kinematics U1-3
Unit 1 Review, pages 100107Knowledge1. (c)
2. (c)
3. (b)
4. (d)
5. (b)6. (c)
7. (d)
8. (b)
9. (d)
10. (b)
11. (b)
12. True
13. True
14. False. The average velocity of an object is the
change in displacementdivided by the change in
time.
15. True16. False. A runner that is veering left to pass
another runnerisaccelerating leftward.
17. False. Since 1987, the annual number of
automobile accident fatalities in Canada has
decreasedby 33%.
18. False. A diagram where 150 m in real life is
represented as 1 cm on the diagram would have a
scale of1 cm : 150 m.
19. False. When given only thex- component and
y-component vectors, the Pythagorean theorem
should be used to determine the magnitude of the
displacement vector.
20. True21. True
22. False. If a bowling ball and a feather are
dropped from the same height in a vacuum at the
same time, then they will both hit the ground at the
same time.
23. True
24. (a) (vii)
(b) (iv)
(c) (v)
(d) (ii)
(e) (iii)
(f) (i)
(g) (vi)25. Answers may vary. Sample answer:
Position is the location of an object relative to a
given reference point. Displacement is the change
in position of an object, while distance is the total
length of the path travelled by an object. For
example, an object that starts at 0 m, moves 20 m
[E] and then 10 m [W] has moved a total distance
of 30 m, but the displacement of the object from its
starting point is only 10 m [E].
26. Answers may vary. Sample answer:
(a) A scalar is a quantity that only has magnitude,
whereas a vector is a quantity that has magnitude
and direction. For example 20 m/s is a scalar since
it has no direction. 20 m/s [E] is a vector since ithas magnitude (20 m/s) and direction (east).
(b) A vector is drawn as a directed line segment,
which is a line segment between two points with
an arrow at one end. The end with the arrow is
called the tip and the end without the arrow is
called the tail. When two vectors are added on a
diagram, one vector is drawn from its starting
point and the second vector is drawn, keeping its
size and direction, but with its tail starting at the
tip of the first vector. The tail of the first vector to
the tip of the second vector is the sum of the two
vectors.
27. Answers may vary. Sample answer:One argument for using speed limiters for teenage
drivers is that teenage drivers are more likely to
speed and cause accidents. Using speed limiters
would lower speeds and decrease the number of
accidents. One of the arguments against using
speed limiters is that by limiting the speed of some
cars they could disrupt the flow of traffic and
cause more problems.
Understanding
28. Given:!
dinital
= 1750 m [W];!
dfinal
= 3250 m [W]
Required: !!
dT
Analysis: !!
dT=
!
dfinal
!
!
dinital
Solution: !!
dT=
!
dfinal
!
!
dinital
= 3250 m [W]! 1750 m [W]
!
!
dT= 1500 m [W]
Statement: My displacement is 1500 m [W].
29. Given:!
dinital
= 2620 m [E];!
dfinal
= 3250 m [W]
Required: !!
dT
Analysis: !!
dT=
!
dfinal
!
!
dinital
Solution: !!
dT =
!
dfinal!
!
dinital
= 3250 m [W]! 2620 m [E]
= 3250 m [W]+ 2620 m [W]
!
!
dT= 5870 m [W]
Statement: My displacement is 5870 m [W].
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Copyright 2011 Nelson Education Ltd. Unit 1: Kinematics U1-4
30. Given:!
dinital
= 1750 m [W];!
dfinal
= 0 m (ignore
the detail about the market because the girls
displacement only involves her initial and final
positions)
Required: !!
dT
Analysis: !!
dT =
!
dfinal!
!
dinital
Solution: !!
dT=
!
dfinal
!
!
dinital
= 0 m !1750 m [W]
= 0 m +1750 m [E]
!
!
dT= 1750 m [E]
Statement: The girls displacement is 1750 m [E].
31. Given:!
dinital
= 121 m [W];!
dfinal
= 64 m [E]
Required: !!
dT
Analysis: !!
dT=
!
dfinal
!
!
dinital
Solution: !!
dT=
!
dfinal
!
!
dinital
= 64 m [E]!121 m [W]
= 64 m [E]+121 m [E]
!
!
dT= 185 m [E]
Statement: The birds displacement is 185 m [E].
32. Given: d= 280 m; t= 4.3 s
Required:vav
Analysis: vav=
!d
! t
Solution: vav=
!d
!t
=
280 m
4.3 s
vav= 65 m/s
Statement: The race cars average speed is
65 m/s.
33. Given: !!
d= 420 m [E]; t= 14.4 s
Required:!
vav
Analysis:!
vav=
!
!
d
! t
Solution:!
vav=
!
!
d
!t
=
420 m [E]
14.4 s!
vav= 29 m/s [E]
Statement: The average velocity of the bird is
29 m/s [E].
34. Given:!
dinital
= 32 km [W];!
dfinal
= 27 km [E];
t= 1.8 h
Required:!
vav
Analysis:!
vav=
!
!
d
!t
!
vav=
!
dfinal
!
!
dinitial
!t
Solution:!
vav=
!
dfinal
!!
dinitial
!t
=27 km [E] ! 32 km [W]
1.8 h
=27 km [E]+ 32 km [E]
1.8 h
=59 km [E]
1.8 h
"
#$
%
&'
1000 m
1 km
"
#$%
&'
=59 000 m [E]
1.8 h
"#$
%&'
1 h
60 min
"
#$%
&'1 min
60 s
"
#$
%
&'
!
vav= 9.1 m/s [E]
Statement: The velocity of the car is 9.1 m/s [E].
35. Given:vav = 263 km/h; t= 13.7 sRequired:d
Analysis: vav=
!d
!t
!d= vav!t
Solution:
!d= vav! t
= 263km
h
!"#
$%&
13.7 s( )1 h
60 min
!
"#$
%&1 min
60 s
!
"#
$
%&
!d= 1.00 km
Statement: The length of the track is 1.00 km or
1000 m.36. (a) The positiontime graph is curved, so the
object has non-uniform velocity. In the first half of
the time, the displacement is more than twice the
displacement in the second half of the time.
(b) The slope at all points of the positiontime
graph is negative, so the velocity is always
negative. The slope is becoming less steep, so the
magnitude of the velocity must also be decreasing.
The object is slowing down as it heads west.
37. (a) Average velocity is the total distance
divided by the total time. Instantaneous velocity is
the velocity at a specific moment. It is possible for
these two values to be different whenever an object
has non-uniform velocity.
(b) Given a positiontime graph, I would calculate
the slope between two points to determine the
average velocity between them. I would look at the
slope of a tangent to the curve to determine the
instantaneous velocity at that point.
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Copyright 2011 Nelson Education Ltd. Unit 1: Kinematics U1-5
38. Given: vi= 0 m/s; t= 1.6 s; vf= 2.8 m/s
Required:aav
Analysis: aav=
!v
!t
aav=
vf! v
i
!t
Solution: aav=
vf !
vi
!t
=
2.8 m/s ! 0 m/s
1.6 s
aav= 1.8 m/s
2
Statement: The average acceleration of the runner
is 1.8 m/s2.
39. Given: vi= 0 m/s; aav = 7.10 m/s2; t= 2.20 s
Required: vf
Analysis: aav=
vf! v
i
!t
aav!t= v
f! v
i
vf= v
i+ a
av!t
Solution:
vf= v
i+ a
av!t
= 0 m/s + 7.10m
s2
!"#
$%&
2.20 s( )!
vf= 15.6 m/s
Statement: The horses final speed is 15.6 m/s.
40. Given:vi = 0 m/s; vf = 152 m/s;
aav = 1.35 104
m/s2
Required:t
Analysis: aav=
vf!
vi
!t
! t=v
f! v
i
aav
Solution: !t=vf! v
i
aav
=
152m
s! 0
m
s
1.35"104m
s2
!t= 1.13"10!2
s
Statement: The arrow will take 1.13 102
s or11.3 ms to accelerate from rest to a speed of
152 m/s.
41. Given:b = 4.0 s; h = 2.0 m/s [E]; l= 4.0 s;
w = 2.0 m/s [E]
Required: !!
d
Analysis: Use the area under the graph to
determine the position at t= 4.0 s:
!
!
d= Atriangle
+Arectangle
Solution:
!
!
d= Atriangle
+Arectangle
= 12bh + lw
=1
24.0 s( ) 2.0
m
s[E]
!"#
$%&+ 4.0 s( ) 2.0
m
s[E]
!"#
$%&
= 4.0 m [E]+ 8.0 m [E]
!
!
d= 12 m [E]
Statement: The object has travelled 12 m [E] after
4.0 s.
42. (a) Given:!
vi= 0 m/s; !
!
d= 15.0 m [down];!
a =!
g = 9.8 m/s2
[down]
Required:t
Analysis: !!
d=!
vi!t+
1
2
!
a !t( )2
= 0 m/s( )! t+1
2
!
a !t( )2
!t( )2
=2!
!
d!
a
!t=2!
!
d!
a
Solution: !t=2!
!
d!
a
=
2 15.0 m( )
9.8m
s2
!"#
$%&
!t= 1.7 s
Statement: The camera takes 1.7 s to hit the
ground.
(b) Given:!
vi= 0 m/s; !
!
d= 10.0 m [down];!
a =!
g = 9.8 m/s2
[down]
Required:!
vf
Analysis: vf
2= v
i
2+ 2a!d
vf= v
i
2+ 2a!d
Solution: vf= v
i
2+ 2a!d
= 0m
s
!"#
$%&
2
+ 2 9.8m
s2
!"#
$%&
15.0 m( )
vf= 17 m/s
Statement: The final velocity of the camera is
17 m/s.
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Copyright 2011 Nelson Education Ltd. Unit 1: Kinematics U1-6
43. (a)
(b)
(c)
44. For each vector, determine the complementary
angle, then reverse the order of the directions.
(a) 90 8 = 82
!
!
d= 86 m [E 8 N]
!
!
d= 86 m [N 82 E]
(b) 90 23 = 67
!
!
d= 97 cm [E 23 S]
!
!
d= 97 cm [S 67 E]
(c) 90 68 = 22
!
!
d= 3190 km [S 68 W]
!
!
d= 3190 km [W 22 S]
45. Given: !!
d1= 850 m [W]; !
!
d2= 1150 m [N]
Required: !!
dT
Analysis: !!
dT= !
!
d1+ !
!
d2
Solution:
This figure shows the given vectors, with the tip of
!
!
d1
joined to the tail of!
!
d2. The resultant vector
!
!
dT
is drawn in black from the tail of !!
d1
to the
tip of !!
d2. Using a compass, the direction of !
!
dT
is [W 54 N]. !!
dT
measures 5.7 cm in length, so
using the scale of 1 cm : 250 m, the actual
magnitude of !!
dT
is 1400 m.
Statement: The students net displacement is
1400 m [W 54 N].
46. Given: !!
d1 = 2.1 m [N];
!
!
d2
= 0.91 m [N 63 E]
Required: !!
dT
Analysis: !!
dT= !
!
d1+ !
!
d2
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Copyright 2011 Nelson Education Ltd. Unit 1: Kinematics U1-7
Solution:
This figure shows the given vectors, with the tip of
!
!
d1
joined to the tail of!
!
d2. The resultant vector
!
!
dT is drawn in black from the tail of!
!
d1 to thetip of
!
!
d2. Using a compass, the direction of !
!
dT
is [N 18 E]. !!
dT
measures 5.2 cm in length, so
using the scale of 1 cm : 0.5 m, the actual
magnitude of !!
dT
is 2.6 m.
Statement: The net displacement of the cue ball is
2.6 m [N 18 E].
47.
dx dy dT
6 8 10
5.0 12 13
8.0 15 172.0 7.0 7.3
6.0 6.7 9.0
Use the Pythagorean theorem to determine teach
missing magnitude: dx2
+ dy2
= dT2.
Row 1:
dT
2= d
x
2+ d
y
2
dT= d
x
2+ d
y
2
= 6( )2
+ 8( )2
dT= 10
Row 2:
dT
2= d
x
2+ d
y
2
dy= d
T
2! d
x
2
= 13( )2
! 5.0( )2
dy= 12
Row 3:
dT
2= d
x
2+ d
y
2
dx= d
T
2! d
y
2
= 17( )2
! 15( )2
dx= 8.0
Row 4:
dT
2= d
x
2+ d
y
2
dy= d
T
2! d
x
2
= 7.3( )2
! 2.0( )2
dy= 7.0
Row 5:
dT
2= d
x
2+ d
y
2
dT= d
x
2+ d
y
2
= 6.0( )2
+ 6.7( )2
dT= 9.0
48.
!
dx
!
dy
5.0 [E] 12.0 [N] [E 67 N]
15.00 [W] 8.00 [N] [W 28 N]
91.0 [E] 151 [S] [E 58.9 S]
640 [W] 213 [N] [W 18.4 N]
0.051 [W] 0.10 [S] [W 63 S]
Use the tangent function: tan!=dy
dx.
Row 1: Find the missing angle.
an!=dy
dx
tan!=12.0
5.0tan!= 2.4
!= tan"1(2.4)
!= 67
Row 2: Find the missing angle.
an!=
dy
dx
tan!=8.00
15.00tan!= 5.33
!= tan"1(5.33)
!= 28
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Row 3: Find the missing component vector.
tan! =dy
dx
tan 58.9 =dy
91.0(1.659)(91.0) = d
y(two extra digits caried)
151 = dy
Row 4: Find the missing component vector.
tan! =dy
dx
tan 18.4 =213
dx
dx
=213
0.33(one extra digit carried)
dx
= 640
Row 5: Find the missing component vector.
tan! =dy
dx
tan 63 =dy
0.051(1.96078)(0.051) = d
y(two extra digits caried)
0.10 = dy
49. (a) Given: !!
dT
= 82 m [W 76 S]
Required:!!
dx
; !!
dy
Analysis: !!
dT= !
!
dx+ !
!
dy
Solution: Since the direction of !!
dT
is between
west and south, the direction of !!
dx
is [W] and the
direction of !!
dy
is [S].
sin! =!d
y
!dT
!dy= !d
Tsin!
= 82 m( ) sin76( )!d
y= 80 m
cos! =!dx
!dT
!dx= !d
Tcos!
= 82 m( ) cos76( )!d
x= 20 m
Statement: The vector has a horizontal or
x-component of 20 m [W] and a vertical or
y-component of 80 m [S].
(b) Given: !!
dT
= 34 m [E 13 N]
Required:!!
dx
; !!
dy
Analysis: !!
dT = !!
dx + !!
dy
Solution: Since the direction of !!
dT
is between
east and north, the direction of !!
dx
is [E] and the
direction of !!
dy
is [N].
sin! =!d
y
!dT
!dy= !d
Tsin!
= 34 m( ) sin13( )!d
y= 7.6 m
cos! =!d
x
!dT
!dx= !d
Tcos!
= 34 m( ) cos13( )!d
x= 33 m
Statement: The vector has a horizontal or
x-component of 33 m [E] and a vertical or
y-component of 7.6 m [N].
(c) Given: !!
dT
= 97 m [S 65 W]
Required:!!
dx ; !
!
dy
Analysis: !!
dT= !
!
dx+ !
!
dy
Solution: Since the direction of !!
dT
is between
south and west, the direction of !!
dx
is [W] and the
direction of !!
dy
is [S].
cos! =!d
x
!dT
!dx= !d
Tcos!
= 97 m( ) cos65( )
!dx = 41 m
sin! =!d
y
!dT
!dy= !d
Tsin!
= 97 m( ) sin65( )!d
y= 88 m
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Statement: The vector has a horizontal or
x-component of 41 m [W] and a vertical or
y-component of 88 m [S].
50. (a) Given: !!
dx
= 4.0 m [W]; !!
dy
= 1.9 m [S]
Required: !!
dT
Analysis: !!
dT = !!
dx + !!
dy
Solution: Let represent the angle !!
dT
makes
with thex-axis.
!
!
dT= !
!
dx+ !
!
dy
!dT
2= !d
x
2+ !d
y
2
!dT= !d
x
2+ !d
y
2
= 4.0 m( )2
+ 1.9 m( )2
!dT= 4.4 m
tan!=!d
y
!dx
tan!=1.9 m
4.0 m
!= 25
Statement: The sum of the two vectors is 4.4 m
[W 25 S].
(b) Given: !!
dx
= 1.9 m [E]; !!
dy
= 7.6 m [N]
Required: !!
dT
Analysis: !!
dT= !
!
dx+ !
!
dy
Solution: Let represent the angle !!
dT
makes
with thex-axis.
!
!
dT= !
!
dx+ !
!
dy
!dT
2= !d
x
2+ !d
y
2
!dT= !d
x
2+ !d
y
2
= 1.9 m( )2
+ 7.6 m( )2
!dT= 7.8 m
tan!=
!dy
!dx
tan!=7.6 m
1.9 m
!= 76
Statement: The sum of the two vectors is 7.8 m
[E 76 N].
(c) Given: !!
dx
= 72 m [W]; !!
dy
= 15 m [N]
Required: !!
dT
Analysis: !!
dT= !
!
dx+ !
!
dy
Solution: Let represent the angle !!
dT
makes
with thex-axis.!
!
dT= !
!
dx+ !
!
dy
!dT
2= !d
x
2+ !d
y
2
!dT= !d
x
2+ !d
y
2
= 72 m( )2
+ 15 m( )2
!dT= 74 m
tan!=!d
y
!dx
tan!= 15 m72 m
!= 12
Statement: The sum of the two vectors is 74 m
[W 12 N].
51. Given: !!
d1= 32 m [W 14 S];
!
!
d2
= 15 m [E 62 S]
Required: !!
dT
Analysis: !!
dT= !
!
d1+ !
!
d2
Solution: Determine the totalx-component and
y-component of!
!
dT :!
dTx
= !
!
d1x+ !
!
d2x
= 32 m( ) cos14( ) [W]+ 15 m( ) cos62( ) [E]= 31.05 m [W]+ 7.04 m [E]
= 31.05 m [W]! 7.04 m [W]
= 24.01 m [W]!
dTx
= 24 m [W]
!
dTy
= !
!
d1y+ !
!
d2y
= 32 m( ) sin14( ) [S]+ 15 m( ) sin62( ) [S]= 7.74 m [S] + 13.24 m [S]
= 20.98 m [S]!
dTy
= 21 m [S]
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Determine the magnitude of !!
dT
:
!dT
2= d
Tx
2+ d
Ty
2
!dT= d
Tx
2+ d
Ty
2
= 24.01 m( )2
+ 20.98 m( )2
(two extra digits carried)
!dT= 32 m
Let represent the angle !!
dT
makes with the
x-axis.
tan!=!d
Ty
!dTx
tan!=20.98 m
24.01 m(two extra digits carried)
!= 41
Statement: The net displacement of the disc is
32 m [W 41 S].52. (a) Given:d= 64 m; v = 0.2 m/s
Required:t
Analysis: v =!d
!t
!t=!d
v
Solution: !t=!d
v
=
64 m
0.2m
s
!t=
3.2 !10
2
s
Statement: It takes the fish 3.2 102
s to cross the
river.
(b) Given:!
v1= 0.90 m/s [S];
!
v2= 0.2 m/s [E]
Required:!
vT
Analysis:!
vT=
!
v1+
!
v2
Solution: Let represent the angle !!
vT
makes
with they-axis.!
vT=
!
v1+
!
v2
vT
2= v
1
2+ v
2
2
vT = v1
2
+ v2
2
= 0.90 m/s( )2
+ 0.2 m/s( )2
vT= 0.9 m/s
tan!=v
2
v1
tan!=0.2 m/s
0.9 m/s
!= 13
Statement: The resulting velocity of the fish is0.9 m/s [S 13 E].
(c) Given:!
vx
= 0.90 m/s [S], t= 3.2 102
s
Required: !!
dx
Analysis:!
vx=
!
!
dx
!t
!
!
dx=
!
vx!t
Solution: !!
dx=
!
vx!t
= 0.90m
s[S]
!"#
$%&
320 s( )
=
288 m [S]!
!
dx= 2.9 '102 m [S]
Statement: The fish arrives 2.9 102
m
downstream from being directly across from where
it started.
53. The pens will hit the ground at the same time.
Both pens have no vertical component to their
initial velocities: the first starts with no velocity at
all and the second starts with only horizontal
velocity. Since both accelerate down at the same
rate (gravity), both pens will land at the same time.
54. Given:vi = 22 m/s; = 62
Required:
!
vix ;!
viy
Analysis:!
vi=
!
vix+
!
viy
Solution: sin! =
!
viy
!
vi
!
viy=
!
visin!
= 22 m/s( ) sin62( )!
viy= 19 m/s
cos! =
!
vix
!
vi
!
vix=
!
vicos!
= 22 m/s( ) cos62( )!
vix= 10 m/s
Statement: The initial velocity has a horizontal or
x-component of 10 m/s and a vertical or
y-component of 19 m/s.
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55. (a) Given:dy = 1.2 m; ay = 9.8 m/s2;
vy = 0 m/s
Required: t
Analysis: !dy= v
y!t+
1
2ay!t
2
!
dy= 0 +
1
2 ay!
t
2
! t2=
2!dy
ay
!t=2!d
y
ay
Solution: !t=2!d
y
ay
=
2 !1.2 m( )
!9.8m
s2
"
#$
%
&'
= 0.4949 s
!t= 0.49 s
Statement: The time of flight of the tennis ball
will be 0.49 s.
(b) Given:ax = 0 m/s2; vx = 5.3 m/s
Required: dx
Analysis: !dx= v
x!t
Solution:
!dx= v
x!t
= 5.3m
s
!
"
#$
%
& 0.4949 s( ) (two extra digits carried)
!dx= 2.6 m
Statement: The range of the tennis ball will be
2.6 m.
Analysis and Application
56. Given:!
vi= 0.60 m/s [up];
!
vf= 27.0 m/s [down]; t= 5.50 s
Required:!
aav
Analysis:!
aav=
!
vf!
!
vi
!t
Solution:!
aav=
!
vf!
!
vi
!t
=27.0 m/s [down]! 0.60 m/s [up]
5.50 s
=27.0 m/s [down]+ 0.60 m/s [down]
5.50 s!
aav= 5.0 m/s
2[down]
Statement: The average acceleration of the roller
coaster is 5.0 m/s2
[down].
57. (a) Given: !!
v = 6.0 m/s [S]; t= 3.0 s
Required:!
aav
Analysis:!
aav=
!!
v
!t
Solution:!
aav=
!!
v
!t
=
6.0 m/s [S]
3.0 s!
aav= 2.0 m/s2 [S]
Statement: The average acceleration from 0 s to
3.0 s is 2.0 m/s2
[S].
(b) Given: !!
v = 6.0 m/s; t= 4.0 s
Required:!
aav
Analysis:!
aav=
!!
v
!t
Solution:!
aav=
!!
v
!t
=
6.0 m/s [S]
4.0 s!
aav= 1.5 m/s2 [S]
Statement: The average acceleration from 2.0 s to
6.0 s is 1.5 m/s2
[S].
(c) Given:b = 5.0 s; h = 10.0 m/s [W]; l= 1.0 s
Required: !!
d
Analysis: !!
d= Atriangle
+Arectangle
Solution:
!
!
d= Atriangle
+Arectangle
=1
2bh + lh
=1
25.0 s( ) 10.0
m
s[S]
!"#
$%&+ 1.0 s( ) 10.0
m
s[S]
!"#
$%&
= 25 m [S]+10 m [S]
!
!
d= 35 m [S]
Statement: The object has travelled 35 m [S] after
6.0 s.
58. Graph (a) shows uniformly increasing velocity.
Graph (b) shows uniformly decreasing velocity.
Every second, the velocity in graph (a) changes by1 m/s while the velocity in graph (b) changes by
1.5 m/s2. So, graph (b) must have the greater
acceleration in magnitude since its velocity is
changing faster.
59. (a) The object has constant positive
acceleration, so the graph will be curved up. Since
the object started from rest at the reference point,
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Copyright 2011 Nelson Education Ltd. Unit 1: Kinematics U1-12
the graph will start at the origin and increase,
curving upward.
(b) The object has constant positive acceleration,
so the graph will be curved up. Since the object
started with negative velocity at the reference
point, the graph will start at the origin, decrease,
level out, then increase up to and beyond the
x-axis.
(c) The object has constant negative acceleration,
so the graph will be curved down. Since the object
started with positive velocity away from the
reference point, the graph will start high on the
y-axis, increase, level out, then decrease down to
and beyond thex-axis.
(d) The object has no acceleration, so the graph
will be a straight line. Since the object started with
negative velocity away from the reference point,
the graph will start high on they-axis, then
decrease down to and beyond thex-axis.
60. (a) Given:vi = 145 km/h; vf= 0 m/s;
a = 10.4 m/s2
Required:t
Analysis: vf= v
i+ a
av! t
!t=v
f! v
i
aav
Solution: Convert vi to metres per second:
vi= 145
km
h
!"# $
%&
1000 m
1 km
!"#
$%&
1 h
60 min
!"#
$%&
1 min
60 s
!"# $
%&
vi= 40.278 m/s
!t=v
f! v
i
aav
=
0 km/h ! 40.278 m/s
!10.4 m/s2
(two extra digits carried)
=
!40.278m
s
!
10.4
m
s 2
!t= 3.87 s
Statement: The car takes 3.87 s to stop.
(b) Given:vf= 0 m/s; a = 11.0 m/s2
Required:d
Analysis: vf
2= v
i
2+ 2a!d
!d=v
f
2! v
i
2
2a
Solution:
!d=v
f
2! v
i
2
2a
=
0 m/s( )2
!40.278 m/s( )
2
2 !10.4 m/s2( )
(two extra digits carried)
=
!1622.3m
2
s2
!20.8m
s2
!d= 78.0 m
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Statement: The car travels 78.0 m while slowing
down.
61. (a) Given:vi = 54.0 km/h; aav = 1.90 m/s2;
t= 7.50 s
Required: !!
d
Analysis:!
!
d=!
vi
!t+1
2
!
aav
!t2
Solution: Convert vi to metres per second:
vi= 54.0
km
h
!
"#
$
%&
1000 m
1 km
!
"#$
%&1 h
60 min
!
"#$
%&1 min
60 s
!
"#
$
%&
vi= 15 m/s
!d= vi!t+
1
2a
av!t
2
= 15m
s
!"#
$%&
7.50 s( ) +1
21.90
m
s2
!
"#
$
%& 7.50 s( )
2
!d= 166 m
Statement: The displacement of the vehicle is
166 m.
(b) Given:vi = 15 m/s; a = 1.90 m/s2; t= 7.50 s
Required:vf
Analysis: vf= v
i+ a!t
Solution:
vf= v
i+ a!t
= 15m
s+ 1.90
m
s2!"# $%&
7.50 s( )
= 29.25m
s
!"#
$%&
1 km
1000 m
!"#
$%&
60 s
1 min
!
"#$
%&60 min
1 h
!
"#
$
%&
!
vf= 105 km/h
Statement: The vans final velocity is 105 km/h.
62.
Equation Uses
Equation 1!
!
d=
!
v1+
!
v2
2
!
"#$
%&! t
Solving for! !d, !v1 , !v2 , ortwhen the other three are knownand acceleration is not known.
Equation 2!
vf=
!
vi+
!
a!t Solving for
!
vi,
!
vf,
!
a , ortwhen the other three are known
and displacement is not known.
Equation 3!
!
d=!
vi!t+
1
2
!
aav!t
2 Solving for! !d, !vi , !a , ortwhen the other three are knownand the final velocity is not known.
Equation 4 !vf
2=
!
vi
2+ 2
!
aav!d Solving for
!
vi,
!
vf,
!
a , or !!
d when the other three are known
and the time is not known.
Equation 5!
!
d=!
vf!t!
1
2
!
aav!t
2 Solving for! !d, !vf , !a , ortwhen the other three are knownand the initial velocity is not known.
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Copyright 2011 Nelson Education Ltd. Unit 1: Kinematics U1-14
63. Given: !!
d1= 240 m [W 11 N];
!
!
d2
= 330 m [N 20 E]; t= 22 s
Determine the displacement of the boat:
Required: !!
dT
Analysis: !!
dT
= !
!
d1
+ !
!
d2
Solution:
This figure shows the given vectors, with the tip of
!
!
d1
joined to the tail of!
!
d2. The resultant vector
!
!
dT
is drawn in black from the tail of !!
d1
to the
tip of!
!
d2. Using a compass, the direction of !
!
dT
is [N 1.4 W]. !!
dT
measures 3.8 cm in length, so
using the scale of 1 cm : 100 m, the actual
magnitude of !!
dT
is 380 m.
Statement: Her displacement is 380 m [W 71 N].
Determine the average velocity of the boat:
Required:!
vav
Analysis:!
vav=
!
!
d
! t
Solution:!
vav=
!
!
d
!t
=
380 m [W 71 N]
22 s
= 17.27 m/s [W 71 N]!
vav=
17 m/s [W 71 N]
Statement: Her average velocity is
17 m/s [W 71 N].
64. Answers may vary. Sample answer:
I would draw what I knew of the vector addition to
determine the magnitude and direction of the
missing vector. For example, I would put the tips
of the two vectors together, and the missing
component vector would be the vector from the tail
of the other component vector to the tip of the
resultant vector.
65. Answers may vary. Sample answer:
(a) Vectors can be added using a scale diagram or
algebraically.
To add vectors using a scale diagram in one or two
dimensions, draw the second vector starting at thetip of the first vector. The distance from the tail of
the first to the tip of the second is the resultant
vector.
Adding vectors algebraically is different
depending on the number of dimensions. With one
dimension, just add the values together once they
have the same direction. In two dimensions, add
thex-components andy-components separately,
then use the tangent function to determine the
direction of the resultant.
(b) Adding vectors using a scale diagram is the
same in one and two dimensions. Using algebra,
you must break the vectors into their components,then solve each dimension by adding as you would
in one dimension.
66. Given:dx = 1250 m; dT = 1550 m
Determine how far north the ranger travelled:
Required: dx
Analysis: !dT
2= !d
x
2+ !d
y
2
!dy
2= !d
T
2! !d
x
2
!dy= !d
T
2! !d
x
2
Solution: !dy= !d
T
2! !d
x
2
= 1550 m( )2
!1250 m( )
2
!dy= 917 m
Statement: The ranger travelled 917 m north.
Determine the direction the ranger travelled:
Required:
Analysis: sin! =!d
x
!dT
Solution: Let represent the angle !!
dT
makes
with they-axis.
sin! =!d
x
!dT
=1250 m
1550 m
! = 53.8
Statement: The ranger travelled in the direction
[N 53.8 E].
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Copyright 2011 Nelson Education Ltd. Unit 1: Kinematics U1-15
67. Given:dx = 11 m; = [N 28 W]
Required: dT
Analysis: sin! =!d
x
!dT
!dT=
!dx
sin!
Solution: !dT=!d
x
sin!
=11 m
sin28
!dT= 23 m
Statement: The archer is 23 m from her target.
68. (a) Given: t= 3.0 s; !!
dx
= 9.0 m [E];!
vy
= 4.0 m/s [S]
Required:!
vT
Analysis:!
vT=
!
vx+
!
vy
Solution: Determine!
vx
, which is constant:
!
vx=
!
!
dx
!t
=
9.0 m [E]
3.0 s!
vx= 3.0 m/s [E]
Use the Pythagorean theorem:
vT
2= v
x
2+ v
y
2
vT= v
x
2+ v
y
2
= 3.0 m/s( )2
+ 4.0 m/s( )2
vT= 5.0 m/s
Let represent the angle!
vT
makes with thex-axis.
tan!=vy
vx
=4.0 m/s
3.0 m/s
!= 53
Statement: The velocity of the object at t= 3.0 s is
5.0 m/s [E 53 S].
(b) Given: t= 3.0 s; !!
dx
= 6.0 m [E];!
ay
= 2.0 m/s2
[S]
Required:!
vT
Analysis:!
vT=
!
vx+
!
vy
Solution: Determine!
vx
, which is constant:
!
vx=
!
!
dx
! t
=
6.0 m [E]
3.0 s!
vx= 2.0 m/s [E]
Determine!
vy
:
!
vy=
!
ay!t
= 2.0m
s 2[S]
!"#
$%&
3.0 s( )!
vy= 6.0 m/s [S]
Use the Pythagorean theorem:
vT
2= v
x
2+ v
y
2
vT= v
x
2+ v
y
2
= 2.0 m/s( )2
+ 6.0 m/s( )2
vT= 6.3 m/s
Let represent the angle!
vT
makes with thex-axis.
tan!=vy
vx
=6.0 m/s
2.0 m/s
!= 72
Statement: The velocity of the object at t= 3.0 s is
6.3 m/s [E 72 S].69. (a) Given:ay = 9.8 m/s
2; vi = 27.5 m/s;
= 41
Determine the time of flight:
Required: t
Analysis: !dy= v
y!t+
1
2ay!t2
Solution:
!dy= v
y!t+
1
2ay!t
2
= vi
sin!( )!t+1
2ay!t
2
0 = 27.5 m/s( ) sin41( )!t+ 12"9.8 m/s2( )!t
2
0 = 18.04 m/s( )!t" 4.9 m/s2( )!t2
0 = 18.04 m/s( ) " 4.9 m/s2( )!t(!t# 0)
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Copyright 2011 Nelson Education Ltd. Unit 1: Kinematics U1-16
!t=
18.04m
s
4.9m
s2
= 3.682 s
!t= 3.7 s
Statement: The footballs time of flight is 3.7 s.
Determine the range:
Required: dx
Analysis: !dx= v
x!t
Solution:
!dx= v
x!t
= vicos!!t
= 27.5m
s
"#$
%&'
cos41( ) 3.682 s( ) (two extra digits carried)
!dx= 76 m
Statement: The footballs range is 76 m.
Determine the maximum height:
Required: dy
Analysis: vfy
2= v
iy
2+ 2a
y!d
y
!dy=
vfy
2! v
iy
2
2ay
Solution: !dy=
vfy
2 ! viy
2
2ay
=0 ! v
isin41( )
2
2 !9.8 m/s2
( )=
0 ! 27.5 m/s( ) sin41( )"# $%2
!19.6 m/s2
=
325.5m
2
s2
19.6m
s2
!dy= 17 m
Statement: The football reached a maximum
height of 17 m.
(b) Given:ay = 9.8 m/s2; t= 3.2 s;
dx = 29 m
Determine the initial velocity:
Required: vi
Analysis:!
vi=
!
vix+
!
viy
Solution: Determine thex-component:
vix=
!dx
!t
=
29 m
3.2 s
= 9.0625 m/s
vix= 9.1 m/s
Determine they-component:
!dy= v
iy!t+
1
2ay!t
2
viy=
!dy!
1
2ay!t
2
!t
=
0 !1
2!9.8 m/s2( ) 3.2 s( )
2
3.2 s
=
4.9
m
s2
"
#$%
&'10.24 s
2
( )3.2 s
= 15.68 m/s
viy= 16 m/s
Use the Pythagorean theorem:
vi
2= v
ix
2+ v
iy
2
vi= v
ix
2+ v
iy
2
= 9.062 m/s( )2
+ 15.68 m/s( )2
(two extra digits carried)
vf = 18 m/s
Statement: The football is kicked with an initial
velocity of 18 m/s.
Determine the initial angle:
Required:
Analysis: tan! =v
iy
vix
tan! =v
iy
vix
=
15.68m
s
9.062m
s
(two extra digits carried)
! = 60
Statement: The football is kicked at an angle of
60.
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Copyright 2011 Nelson Education Ltd. Unit 1: Kinematics U1-17
70. (a) If two objects were dropped at the same
time from the same height, but one was on the
Moon and the other here on Earth, then the one
that is dropped on Earth will hit the ground first
since the acceleration due to Earths gravity is
greater than the acceleration due to the Moons
gravity.(b) If one beanbag was launched horizontally and
another of equal mass was dropped from the same
height at the same time, then they would both hit
the ground at the same time. If this experiment
were performed on the Moon the results would be
the same since both beanbags would still
experience the same vertical acceleration.
(c) For the beanbags launched horizontally in part
(b), the beanbag on the Moon would have the
larger range. Since both beanbags are launched
with the same horizontal velocity, the one with the
larger time of flight will have the larger range.
Since the Moon has less gravity, the beanbag onthe Moon will fall more slowly and have a longer
time of flight than the one on Earth.
71. Suppose a projectile in a parabolic path was
surrounded by a room. Within that room, the
projectile would appear to be free-floating without
gravity. So, by flying an airplane in the parabolic
path of a projectile, the people in the airplane will
experience what feels like a gravity-free
environment inside the airplane.
Evaluation
72. Answers may vary. Sample answer:(a) Drawing vectors in three dimensions would use
the same directed line segments, except that
instead of only pointing in directions on a surface,
they would point in any direction in a space.
(b) Adding two vectors on a three dimensional
diagram would apply the same methods used for
one and two dimensions. The first vector would be
drawn and the second added by joining its tail to
the tip of the first vector. The resultant vector
would then be the vector drawn from the tail of the
first to the tip of the second.
(c) Three dimensional vectors would have three
component vectors. These components wouldcorrespond to the normalx- andy-axes vectors:
east and west for thex-directions, north and south
for they-axis vectors, and then there would be an
additional direction on a verticalz-axis
corresponding to up and down.
73. Answers may vary. Sample answer:
Yes, Galileos experiments had a major impact and
lead to the discovery of modern kinematics. Not
only did the experiments lead to important
scientific discoveries that enabled Newton to
formulate his theories, but he was also one of the
first to challenge the scientific notions of the day.
Without his leadership, people may have been left
with their false understandings for another century.
74. Answers may vary. Sample answer:Accelerometers could be used in wireless mice to
adjust the power needed for the signal. Wireless
mice already use blue tooth technology to allow
you to control your computer remotely, but this is
usually limited to a short distance. Accelerometers
could be used in the mouse to measure its position
from the receiver. If the distance increases, the
mouse could draw more power. If the distance
decreases, this could also be used as a power
saving technology because the mouse is close to
the receiver and does not need as strong of a
signal.
Reflect on Your Learning75. Answers may vary.
(a) Students should discuss any material in the unit
that they found illuminating or insightful.
(b) Students should discuss their understanding of
trigonometry and its application in direction and
projectiles.
76. Answers may vary. Students should discuss
their understanding of gravity and projectiles.
Research
77. Answers may vary. Students answers shoulddiscuss the use of high-speed rails around the
world. They may wish to include which countries
have the most high-speed trains, and which
countries have the fastest trains. They will
probably discuss Canadas proposed high-speed
rail locations and the issues involved.
78. Answers may vary. Students answers should
discuss a sport and the world record speeds and
distances involved. They should also include their
group work and calculations performed to
determine how their values compare with the
record values.
79. Answers may vary. Students answers shoulddiscuss the future of space travel and any new
technologies that may make it more easily
accessible. Topics include but are not limited to
space elevators, warp drives, the future of private
space flight, and designs for space shuttles that
carry their own fuel.